The smallest angle of the triangle whose side | vedclass

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(B) Let the sides of the triangle be $a = 6+\sqrt{12} = 6+2\sqrt{3}$,$b = \sqrt{48} = 4\sqrt{3}$,and $c = \sqrt{24} = 2\sqrt{6}$.Note that $a \approx

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$\frac{\pi}{6}$

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(B) Let the sides of the triangle be $a = 6+\sqrt{12} = 6+2\sqrt{3}$,$b = \sqrt{48} = 4\sqrt{3}$,and $c = \sqrt{24} = 2\sqrt{6}$.Note that $a \approx 6 + 3.464 = 9.464$,$b \approx 4 	imes 1.732 = 6.928$,and $c \approx 2 	imes 2.449 = 4.898$.The smallest side is $c = 2\sqrt{6}$. The smallest angle is opposite to the smallest side,so we need to find angle $C$.Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.$a^2 = (6+2\sqrt{3})^2 = 36 + 12 + 24\sqrt{3} = 48 + 24\sqrt{3}$.$b^2 = (4\sqrt{3})^2 = 48$.$c^2 = (2\sqrt{6})^2 = 24$.$\cos C = \frac{48 + 24\sqrt{3} + 48 - 24}{2(6+2\sqrt{3})(4\sqrt{3})} = \frac{72 + 24\sqrt{3}}{8\sqrt{3}(6+2\sqrt{3})} = \frac{24(3 + \sqrt{3})}{8\sqrt{3} 	imes 2(3 + \sqrt{3})} = \frac{24}{16\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.Since $\cos C = \frac{\sqrt{3}}{2}$,we have $C = \frac{\pi}{6}$.

The smallest angle of the triangle whose sides are $6+\sqrt{12}, \sqrt{48}, \sqrt{24}$ is

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